Enthalpy change in reaction?
10mL 0.5M Na2CO3 + 5mL 1M CaCl2?
Ccal is 0.05326 J/deg celsius
temperature change is -1 deg celsius
CaCO3 specific heat is 0.8181 J/g.deg celsius
Answer:
The reaction can be written as
Na2CO3 + CaCl2 ------> 2NaCl + CaCO3
Thus, one mole each of Na2CO3 and CaCl2 react to produce one mole of CaCO3.
Here we have (0.010*0.5)=5*10^-3 moles of Na2CO3 and (0.005*0.1)=5*10^-3 moles of CaCl2
So we have 5*10^-3 moles of CaCO3. That is 0.5 grams of CaCO3 produced in the reaction. (The molecular weight of CaCO3 is 100)
The total heat exchanged in the reaction = Heat lost by (Calorimeter + CaCO3) = (0.05326*1)+(0.8181*0.5*1) = 0.4623 Joules
This is the change in enthalpy for 5*10^-3 moles
The standard enthalpy change is defined for 1 mole
So, the standard enthalpy change = 0.4623/(5*10^-3) = 92.462 Joules/mole
(We have the enthalpy positive as the reaction is an endothermic one)
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